Simultaneity: Train and Lightning Thought Experiment

[1977ub]

Suppose a star five light years away explodes. Eventually the light from the explosion reaches your eyes and you observe it. When did the star explode? The only sensible answer is "five years before the light reached your eyes" because it took five years for the light to travel from the explosion to your eyes.

This is "sensible" in the IF of the observer. For someone following the light from the explosion at near light speed, the answer is "almost instantly before the earther saw it" - of course he would synchronize Earth clocks with the exploding star's clock differently than the earther would.

 

[sdkfz]

Here’s my attempt at answering taenyfan’s very fine question on Einstein’s train experiment:

The problem defines two frames of reference moving with respect to one another. In the platform observer’s frame the two strikes are simultaneous. In this scenario the woman must be considered to be moving relative to the lightning strikes, specifically, toward the forward strike and away from the rear strike.

We could just as well consider the train to be “at rest” and the landscape, including the platform observer moving from the train’s front to its rear. In this case the woman will see the two strikes as simultaneous but the man — who is “moving” toward the back of the train and away from the front — will see the rear strike first.

Since the two frames are moving relatively you must pick one or the other when describing the two strikes as “simultaneous”. You can’t have it both ways.

Worded like this, there's a danger that it might make it seem like the same two strikes are simultaneous, for both observers, and which observer considers them to be simultaneous is a choice.

It's absolutely true that two strikes could occur that happen to be simultaneous for the train observer; but these two strikes can't be simultaneous for the platform observer, no matter what frame of reference is used.

(I'm sure that's what you meant, I was just worried about how the bit I bolded could be read by someone - that it's the same two strikes that the standard thought experiment stipulates are simultaneous in the embankment frame. I'd have worded it like "In this case the woman might see two other strikes as simultaneous but the man ...".)

 

[Nugatory]

This is "sensible" in the IF of the observer. For someone following the light from the explosion at near light speed, the answer is "almost instantly before the earther saw it" - of course he would synchronize Earth clocks with the exploding star's clock differently than the earther would.

That definition of when an observed event happened (time of observation minus light travel time to point of observation) works for all inertial observers - no other definition is "sensible" in the sense of being free of internal inconsistency. The point of the simultaneity thought experiment is that it produces different results for observers in motion relative to one another.

 

[Mister T]

The problem defines two frames of reference moving with respect to one another. In the platform observer’s frame the two strikes are simultaneous. In this scenario the woman must be considered to be moving relative to the lightning strikes, specifically, toward the forward strike and away from the rear strike.

In more carefully crafted versions of this scenario the lightning strikes leave char marks on both the train car and the train tracks. Thus the woman on the train is always midway between the char marks on the train and the man on the platform is always midway between char marks on the track. It's the light that was emitted from the lightning strikes that arrive at the observers, not the strikes themselves.

All that matters is that one observer (in this case the woman, but in other versions it could be the man) sees one flash arrive before the other, but the other doesn't.

We could just as well consider the train to be “at rest” and the landscape, including the platform observer moving from the train’s front to its rear. In this case the woman will see the two strikes as simultaneous but the man — who is “moving” toward the back of the train and away from the front — will see the rear strike first.

But you would have to create a different version of the scenario, as mentioned above, not simply change your consideration of who's at rest. In other words, if the flashes arrive at the man at the same time, that is a single event and as such it's invariant. The woman will agree that the flashes arrived at the man at the same time. And she will conclude that the reason that happened is that one of the flashes was emitted before the other.

 

[1977ub]

That definition of when an observed event happened (time of observation minus light travel time to point of observation) works for all inertial observers - no other definition is "sensible" in the sense of being free of internal inconsistency. The point of the simultaneity thought experiment is that it produces different results for observers in motion relative to one another.

Ok so it is sensible in a circular way I guess. "Light travel time" will be different for different observers. When I see a star explode on earth, in the IF which roughly contains the Earth and the star I will conclude the star is pretty consistently 4 LY away and light took 4 Y to get here.

 

[Nugatory]

When I see a star explode on earth, in the IF which roughly contains the Earth and the star I will conclude the star is pretty consistently 4 LY away and light took 4 Y to get here.

However, if you're thinking about it that way you're setting yourself up for future confusion. When you say "the IF which roughly contains the Earth and the star", you seem to be suggesting that there might be an inertial frame (or non-inertial frame, for that matter) that does not "contain" the Earth and the star. That's not right, because everything is always in all frames always - there's no such thing as being in one frame and not another.

There is such a thing as being at rest in one frame and not another, but that doesn't change any of the physics. In particular, the basic principle that the time something happened is the time at which it could be observed minus the light travel time between the event and the point of observation works in all inertial frames. Whether the source or the observer is at rest in the frame is irrelevant.

 

[1977ub]

However, if you're thinking about it that way you're setting yourself up for future confusion. When you say "the IF which roughly contains the Earth and the star", you seem to be suggesting that there might be an inertial frame (or non-inertial frame, for that matter) that does not "contain" the Earth and the star. That's not right, because everything is always in all frames always - there's no such thing as being in one frame and not another.

There is such a thing as being at rest in one frame and not another, but that doesn't change any of the physics. In particular, the basic principle that the time something happened is the time at which it could be observed minus the light travel time between the event and the point of observation works in all inertial frames. Whether the source or the observer is at rest in the frame is irrelevant.

I meant the IF in which the Earth and the star are basically at rest. In this frame, the star explodes and the time it takes for that light to reach Earth is 4 Y. In a different IF - traveling .999c, arriving at the star as it explodes, then passing the Earth - the time between explosion and seeing on Earth is miniscule.

 

[Nugatory]

In a different IF - traveling .999c, arriving at the star as it explodes, then passing the Earth - the time between explosion and seeing on Earth is miniscule.

As is the distance between the two events.

 

[Mister T]

I meant the IF in which the Earth and the star are basically at rest.

Yes, your meaning was clear from the context. That's usually referred to as the rest frame of those objects.

I think the point @Nugatory makes is that using the language one way signals an understanding, but using it another way signals, if not a misconception, then at least a warning that it might lead to a misconception.

In this frame, the star explodes and the time it takes for that light to reach Earth is 4 Y. In a different IF - traveling .999c, arriving at the star as it explodes, then passing the Earth - the time between explosion and seeing on Earth is miniscule.

And calculations done using that frame would also involve the delay due to light travel time.

 

[Josh_Seedman]

No. Assuming the train has Born rigidity (it maintains its proper length in each momentary rest frame during its acceleration), then the clock at the front of the train will run faster than the clock at the back during the acceleration, and when acceleration ends, both clocks will again run at the same speed but will be out of sync in the train's rest frame. In the platform rest frame, both clocks would slow down as the train accelerates at the same rate and remain synchronized. This is due to the fact that the plane of simultaneity at each point on the train is shifting during acceleration, causing "now" for any point on the train ahead of the point of interest (in the direction of acceleration) to shift into the "future" from where it was, and causing "now" for any point behind the point of interest to shift into what was the past of where it was.

The most important fact to remember about the relativity of simultaneity is that if you have two clocks separated by some nonzero distance in space, then they can be synchronized in at most one valid inertial frame of reference. In all others they will not be.

OK. How about I back-peddle by a few steps and instead request the following assumption: The two clocks on the moving train are synchronized only after the train has reached its constant velocity. The sense I get is that it can be agreed that one or more valid procedures exist by which these two clocks can be synchronized to each other, in the IF of the moving train. Yes?

 

[Mister T]

OK. How about I back-peddle by a few steps and instead request the following assumption: The two clocks on the moving train are synchronized only after the train has reached its constant velocity. The sense I get is that it can be agreed that one or more valid procedures exist by which these two clocks can be synchronized to each other, in the IF of the moving train. Yes?

Yes, assuming you mean rest frame of the moving train when you say IF of the moving train.

But then they won't be synchronized in the rest frame of the platform. That's the point.

 

[Ziang]

I found the thought experiment here http://www.bartleby.com/173/9.html
and Einstein concluded:" Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity)."

Now let me place a long seesaw on the embankment and parallel with the railway. This seesaw is working. I mean its both ends A & B are moving up and down continuously.
Let us consider two events: A is at high position and B is at low position. According to the relativity of simultaneity, these two events are simultaneous with reference to the embankment but are not simultaneous with respect to the train. Hence, the woman on the train claims that the seesaw is not straight up but it is curved up and down on its both ends A & B continuously. The lady questions why the seesaw is not broken?

 

[Ibix]

The lady questions why the seesaw is not broken?

Rotating objects generally look rather odd in relativity - you can't have rigid objects, so it's not that the seesaw is broken but rather that it is not rigid. A related phenomenon is the appearance of a wheel at relativistic speed. Even if the wheel is not itself rotating relativistically then it is length contracted into an ellipse as seen from a moving frame, yet it still rotates.

 

[Ibix]

An important point to bear in mind, which I forgot to mention above, is that the seesaw is not straight even in the embankment frame. When one end strikes the ground a mechanical wave propagates up the beam at the speed of sound in the material and the other end doesn't stop rising until that wave reaches it. That's typically on a timescale of milliseconds, which is why you don't notice, but it's always going to be flexing.

Relativity forces you to pay attention to such details. Unfortunately that means that a formal analysis of this problem requires a detailed mechanical model of the seesaw and its reaction to applied forces.

 

[Peter Martin]

The problem is in the description of the situation, which states that the lightning strikes are "simultaneous" without stating that the simultaneity is from the man's (bystander's) point of view. Let's re-describe the problem from the woman's (passenger's) point of view.

A woman sits at the middle of a train. Out the window she sees the countryside - which includes a man standing watching the train - rushing by in the direction of the rear of the train. Suddenly she sees lightning strike the front and rear cars of the train simultaneously. The question is: What does the man see?

Since he is rushing toward the rear of the train, he sees the lightning strike the rear car first because, as the light travels toward him, he is traveling toward the source of the light. By the same token, he is moving away from the strike on the front car so it takes longer for the light to reach him.

Since we typically spend more time on the landscape than on trains, we naturally take the man's point of view when describing this apparent paradox. So as soon as you read the (biased) description you are already on the "wrong track".

 

[PeroK]

I found the thought experiment here http://www.bartleby.com/173/9.html
and Einstein concluded:" Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity)."

Now let me place a long seesaw on the embankment and parallel with the railway. This seesaw is working. I mean its both ends A & B are moving up and down continuously.
Let us consider two events: A is at high position and B is at low position. According to the relativity of simultaneity, these two events are simultaneous with reference to the embankment but are not simultaneous with respect to the train. Hence, the woman on the train claims that the seesaw is not straight up but it is curved up and down on its both ends A & B continuously. The lady questions why the seesaw is not broken?

It's also a question of by how much the seesaw is out of sync. In the frame of the train, clocks at either end of the seesaw will be out of sync by $\frac{Lv}{c^2}$, where $L$ is the rest length of the seesaw and $v$ is the speed of the train. The train must be traveling at less than $c$, so an upper limit on this is $\frac{L}{c}$.

Now, for a seesaw of even $100m$, say, this is a very small time difference, less than a micro-second. The woman on the train will still measure the seesaw as being essentially in sync and observe nothing unusual.

Note that the seesaw will be curved in the platform frame as well, due to the forces along its length.

 

[Ziang]

An important point to bear in mind, which I forgot to mention above, is that the seesaw is not straight even in the embankment frame. When one end strikes the ground a mechanical wave propagates up the beam at the speed of sound in the material and the other end doesn't stop rising until that wave reaches it. That's typically on a timescale of milliseconds, which is why you don't notice, but it's always going to be flexing.

We don't have to use a seesaw moving up and down. Let us use a horizontal seesaw that its two ends move closer and farther from the railway. I mean the horizontal seesaw rotates freely an small angle during the experiment.

 

[Ziang]

Note that the seesaw will be curved in the platform frame as well, due to the forces along its length.

You can imagine a horizontal seesaw. And the seesaw rotates an angle freely during the experiment.

 

[pervect]

I found the thought experiment here http://www.bartleby.com/173/9.html
and Einstein concluded:" Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity)."

Now let me place a long seesaw on the embankment and parallel with the railway. This seesaw is working. I mean its both ends A & B are moving up and down continuously.
Let us consider two events: A is at high position and B is at low position. According to the relativity of simultaneity, these two events are simultaneous with reference to the embankment but are not simultaneous with respect to the train. Hence, the woman on the train claims that the seesaw is not straight up but it is curved up and down on its both ends A & B continuously. The lady questions why the seesaw is not broken?

I would tend to believe the conclusion that the see-saw is curved, as "rigid objects" simply aren't compatible with special relativity. One can define Born-rigid motions in special relativity, but objects satisfying the necessary criterion to be Born-rigid can't change their state of rotation. Your see-saw is changing it's state of rotation, so it can't be Born-rigid.

You may not be familiar with Born rigidity. I don't see how you can learn about it before you learn about the relativity of simultaneity, though. So the process of learning special relativity involves first realizing that simultaneity is relative, then exploring all the logical consequences of this fact (along with the other aspects of SR such as length contraction and time dilation, though the relativity of simultaneity seems to be the hardest thing for people to learn). The lack of rigid objects is one of the logical consequences of special relativity, I'm unsure if it can be formally deduced solely from the relativity of simultaneity however.

I don't think your exposition isn't quite complete, a drawing of the seesaw from the perspective of the ground and from the train using the Lorentz transform would be interesting. I believe your conclusion is probably right, but the argument isn't quite rock solid yet.

 

[Janus]

We don't have to use a seesaw moving up and down. Let us use a horizontal seesaw that its two ends move closer and farther from the railway. I mean the horizontal seesaw rotates freely an small angle during the experiment.

This doesn't change anything about any of the arguments already made. For the seesaw to swing back and forth on any axis, some force must be applied to it at some point, and that force cannot propagate through the seesaw faster than c.

 

[Ziang]

This doesn't change anything about any of the arguments already made. For the seesaw to swing back and forth on any axis, some force must be applied to it at some point, and that force cannot propagate through the seesaw faster than c.

Let me say the seesaw is rotating at a constant angular speed (like the Earth is spinning).
At the time point that the train is passing it, it is parallel with the railway (and is still rotating)

 

[Ibix]

Let me say the seesaw is rotating at a constant angular speed (like the Earth is spinning).
At the time point that the train is passing it, it is parallel with the railway (and is still rotating)

The seesaw will not appear straight. Also its length will vary as it rotates. You may wish to Google for "relativistic wheel" and look at the shapes of the spokes.

 

[Janus]

To build on Ibix's point, Let's consider the following two illustrations:

On the left we have our rotating pole (light blue bar), The tracks (black and brown lines), the train( green line) and our observers on the embankment and on the train (red circles), according to the embankment frame. This is the moment the observers pass each other. The pole is parallel to the tracks, and its ends line up with both the end of the train and particular points on the track (the white lines). The pole is rotating counter-clockwise as shown by the blue arrows.

One thing needs to be noted in this image, the train, since it is moving relative to this frame is length contracted. In other words, the length of the train as measured in this frame is shorter than the length of the train as measured in its own frame.

This becomes apparent when we look at the right image, which is drawn from the frame of the train. We are still dealing with the moment the two observers pass each other. In this frame the train measures its length as its proper length, while it measures the tracks and embankment, which is moving relative to it as being length contracted. As a result, the train no longer fits neatly between the white lines, but extends quite a bit beyond is both directions. This also means that one end ( the right end in this case) has already passed its white line before the other end hasn't reached its white line yet.

Also note that the bar appears as being curved, as per Ibex's post, and the right end of the pole has already gone past the point where it is adjacent to the tracks, while the left end has not yet reached that point. If this were an animation that we could run backwards and forwards, you would see the ends of the train lining up with a white line at the same moment that an end of the rod was adjacent to the same spot. This just happens at different times for each white line. Another thing to note here is that as the rod rotates in this frame, the curvature of the rod does not remain constant but changes from being curved as shown here to being straight when aligned vertically.

 

[Ziang]

The seesaw will not appear straight.

That is what I said: "the woman on the train claims that the seesaw is not straight up but it is curved up and down on its both ends A & B continuously. The lady questions why the seesaw is not broken?"

 

[Janus]

That is what I said: "the woman on the train claims that the seesaw is not straight up but it is curved up and down on its both ends A & B continuously. The lady questions why the seesaw is not broken?"

Just because the See-saw is curved in her frame does not mean that it is under stress in her frame.
In the following image we see the same diamond shape, in both its rest frame and according to a frame in which it is moving at 0.8c.
In the top image all the corner angles are 90 degrees, in the bottom image, one pair of corners is more than 90 degrees and the other pair is less than 90 degrees. But this does not mean that the shape is under some type of stress at the corner. Even if the shape was rotating, and thus constantly changing shape in the bottom frame, it would not be undergoing any stress.

 

[Mister T]

Rotating objects generally look rather odd in relativity

Yes, but this is not about how the seesaw appears, it's about its shape. Observations differ in different frames. The shape of the seesaw is one such observation.

 

[Mister T]

The lady questions why the seesaw is not broken?

Because in its rest frame it's not bent (enough for it to break).

 

[Ibix]

That is what I said: "the woman on the train claims that the seesaw is not straight up but it is curved up and down on its both ends A & B continuously. The lady questions why the seesaw is not broken?"

That's what a freely rotating beam looks like when viewed from a moving frame. It can't be straight - as you've already observed the relativity of simultaneity shows that immediately. Any other maths you do will support this. For example you can transform the centripetal force to show that it's not centripetal in the moving frame, and its direction depends on the radius, which is consistent with a curved beam. Or you can draw a 2+1 dimensional Minkowski diagram and consider the intersection of a sloped simultaneity plane with the helical worldtube of the spinning beam.

The question is, why do you think it ought to be straight? The short answer is that you are trying to use common sense in an extremely uncommon situation. That doesn't work well.

 

[Ziang]

Just because the See-saw is curved in her frame does not mean that it is under stress in her frame.

The laws of physics are invariant in all inertial frames.
If the seesaw gets stress when it gets curved in one inertial frames then it would get stress when it gets curved in all inertial frames.

 

[pervect]

The laws of physics are invariant in all inertial frames.
If the seesaw gets stress when it gets curved in one inertial frames then it would get stress when it gets curved in all inertial frames.

The laws of physics are invariant in all inertial frames. However, "being curved", which I interpret as a statement about the purely spatial geometry of the object, is a frame dependent statement, due to the relativity of simultaneity. If parts of an object have a non-zero proper acceleration - such as a rotating stick, the see-saw, or an accelerating elevator - the spatial projection of the object may be flat in some frames of reference and "curved" in other frames of reference.

Note that if an object is not accelrating, if all parts of the object have a zero proper acceleration, then the property of "being flat" does turn out to be independent of the chocie of frame, as the Lorentz transformation is linear, and linear functions map straight lines to straight lines.

However, the worldline of an accelerating point is not a straight line, so this argument does not apply.

 

[Mister T]

The laws of physics are invariant in all inertial frames.
If the seesaw gets stress when it gets curved in one inertial frames then it would get stress when it gets curved in all inertial frames.

Okay. And if the stress is not enough to break it in one frame, then it doesn't break in all frames. It will be bent by different amounts in different frames, though.

 

[pervect]

The way I would put the stress situation is that the stress-energy tensor is a rank 2 tensor, and transforms as such. I do have some questions on the relationship between the physicists usage of the stress energy tensor, which I'm familiar with, and engineering usage, which I'm not familiar with and which may be slightly different.

I'm tempted to say that if the stress-energy tensor is zero in one frame, it's zero in all frames. While this is a correct statement, it's misleading, because the energy part of the stress-energy tensor isn't zero if one has matter present. So in the cases under consideration, one would actually need to carry out the appropriate transformations to figure out how it transforms.

The ability to handle rank 2 tensors and their transformations comes well after one learns introductory special relativity, however. If one doesn't learn the basics of SR, one will never get this far.

 

[pervect]

I did work out the shape of a rotating bar in a moving reference frame via the Lorentz transform. This is simpler than the see-saw case to analyze, but one can gain some insight of the see-saw case from the rotating bar case.

It's convenient to write the coordinate of any point on the bar as a function of r, which we set to zero at the origin of the bar (the origin being the point on the bar with zero proper acceleration). We will call the inertial frame of reference in which the origin of the bar is at rest "the rest frame of the bar", even though only the origin of the bar is at rest in this frame. r is a parameter that picks out a specific point on the bar. We also need a time coordinate, which we will take as the proper time $\tau$ of the point we just specified by specifying r. We will let the angular frequency of the rotation be $\omega$ in the bar's rest frame. Then in the bar's rest frame, we write:

$$x = r \cos \frac{\omega \tau}{\sqrt{1-r^2\omega^2}} \quad y = r \sin \frac{\omega \tau}{\sqrt{1-r^2\omega^2}} \quad t = \frac{\tau}{\sqrt{1-r^2\omega^2}}$$

Applying the Lorentz transform, we can transform these coordinates to a moving frame. Let the velocity be determined by the dimensionless parameter $\beta$, so that v = \beta c. And let $\gamma = 1/\sqrt{1-\beta^2}$. Then we can find the position of a point on the bar in a moving frame with coordinates x1, y1, t1 via the Lorentz transform.

$$x1 = \gamma \left(x + \beta\,c\,t \right) \quad y1=y \quad t1 = \gamma \left( t +\beta x/c \right) $$

We can substitute the expressions for $x(r,\tau)$, $y(r,\tau)$, and $t(r,\tau)$ from the first set of equations into the second to find $x1(r, \tau), y1(r,\tau), t1(r,\tau)$. It's also convenient to set c=1 at this point, unless one really wants to keep tract of it (I did not).

We now wish to find and plot x1 and y1 as a function of t1, rather than as a function of $\tau$. To do this, we need to solve the equation

$$
t1 = \frac{1}{\sqrt{1-\beta^2}} \left( {\frac {\tau}{\sqrt {1-{r}^{2}{\omega}^{2}}}}+\frac{\beta}{c}\,r\cos
\left( {\frac {\omega\,\tau}{\sqrt {1-{r}^{2}{\omega}^{2}}}} \right)
\right)$$

for $\tau$ as a function of r and t1. This is not something that has a closed form solution. We will call this $\tau = f^{-1}(r, t1)$

Then we can write the coordinates x1 and y1 of the bar at time t1 as a function of r, where I have now omitted the factors of c:

$$ x1 =
\left( r\cos \left( {\frac {\omega\,f^{-1}}{\sqrt {1-{r}^{2}{\omega}^{2}}}} \right) +{\frac {\beta\,f^{-1}}{\sqrt {1-{r}^{2}{\omega}^{2}}}} \right) {
\frac {1}{\sqrt {1-{\beta}^{2}}}}$$

$$ y1=r\sin \left( {\frac {\omega\,f^{-1}}{\sqrt {1-{r}^{2}{\omega}^{2}}}} \right) $$

This would be quite messy to carry out by hand - I used a computer algebra package to do the algebraic manipulations and to do some graphs for $\beta = .9$ and $\omega = .5$ with r varying from 0 to 1. While I don't have a convenient way of posting the results, I can say that the bar is generally not straight (there's an exception when it's vertical), that one can see the effects of time dilation (the bar rotates more slowly), and that the bar is length contracted as well as bent when it's not vertical.

As I mentioned in a previous post, this is more or less to be expected. While the Lorentz transformation is a linear transformation and must map straight lines into straight lines, the worldlines of points on the bar (a fixed value of r as $\tau$ varies) are not straight lines for r>0. So we don't expect the bar to be straight, and we don't expect the bar to have a constant length, and the calculations demonstrate this.

I like this approach because a) I'm used to it, and b) the coordinates r and $\tau$ have physical significance, being the proper distance of a point on the bar from the origin in the former case, and being the proper time of a point on the bar in the later case. Both the proper distance and the proper times are independent of any coordinate choices, so it's a convenient representation of the spinning bar.

 

[Mister T]

Nice analysis @pervect

Suppose the bar is made of some delicate material such as glass so that a very small bend will cause it to break. What the OP is asking about, I believe, is why it doesn't break when it bends in the moving frame. This is just one of those things that's part of special relativity, like time dilation and length contraction. It's like asking how, when a moving rod is Lorentz-contracted, it can withstand the compression without breaking. Relativity contradicts common sense in many ways, and this is just one of those ways. Understanding them expands common sense.

 

[Ibix]

This might be another way to look at it.

In the rest frame of its centre of mass it's straight, and rotated at each subsequent time. So a rotating rod can be viewed as a helical worldsheet. This can be illustrated with a (2+1)d Minkowski diagram - I've drawn one below. One spatial plane of the rest frame is marked as a blue plane (time is perpendicular to this), and naturally the intersection of the green helical worldsheet and the plane is a straight line. So the rod is straight in this frame.

But what happens in a moving frame? In a moving frame, the spatial plane is tilted compared to the one drawn above. That looks like this:

Now you can clearly see that the intersection of the helix and the plane is a curve. That's why the rod is curved in this frame (although as @pervect notes, there's a special case when the rod is "across" the slope where it'll be straight). You can also see immediately that I've changed nothing about the rod by changing the frame. So it can't be broken; it's just the intuitive notion that "if it isn't straight it ought to be broken" that doesn't work properly.